14.(本题4分)-|||-一半径 _(1)=5cm 的金属球A,带电荷 _(1)=+2.0times (10)^-8C, 另一内半径为 _(2)=10cm-|||-外半径为 _(3)=15cm 的金属球壳B,带电荷 _(2)=+4.0times (10)^-8C, 两球同心放置,如图 6-9 所-|||-示.若以无穷远处为电势零点,则A球电势 _(A)= __ B球电势 (I)_(B)=-|||-__ (dfrac (1)(4pi {varepsilon )_(0)}=9times (10)^9dfrac (Ncdot {m)^2}({C)^2})-|||-B-|||-r2-|||-A-|||-0-|||-r-|||-图 6-9